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Math Help - analysis-bounded monotone sequence problem

  1. #1
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    Unhappy analysis-bounded monotone sequence problem

    Prove that the sequence
    lim n-->infinity 1/n(1 + 1/2 + ...+ 1/n) = 0

    is
    (i) monotone
    (ii) bounded
    (iii) find its limits

    How can i proceed to prove that it is monotone and bounded???
    Help me please!
    Thank You!
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  2. #2
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    Is 1 + 1/2 + ... + 1/n in the denominator?

    If it is in the numerator, then this is divergent, since it is greater than 1/n, which is divergent.

    (EDIT: nevermind this. I thought it was a sum!)
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  3. #3
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    If 1 + 1/2 + ... + 1/n is in the denominator, then the denominator itself is increasing (at each step the sum gets bigger, and it gets multiplier by a greater number). This implies that the sequence is decreasing. It is bounded below since all its terms are positive. And it will converge to 0, since it is smaller than 1/n, which converges to 0.
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  4. #4
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    Thank You.
    It is 1/n * (1 + ... )

    How about the monotone?
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  5. #5
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    Thank You.
    It is 1/n * (1 + ... )

    How about the monotone?
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  6. #6
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    an = (1/n) * (1 + 1/2 + ... + 1/n) < (1 + 1/4 + ... +1/n^2)

    this last sequence is convergent since it is the series with term 1/k^2. This proves that your sequence is bounded.
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  7. #7
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    a_n = (1/n) * (1 + 1/2 + ... + 1/n)

    a_(n+1) = 1/(n+1) * (n*a_n + 1/(n+1))

    (n+1) * ( a_(n+1) - a_n ) = 1/(n+1) - a_n = 1/(n+1) - (1/n) * (1 + 1/2 + ... +1/n) < 1/(n+1) - 1/n < 0

    This proves that the sequence is monotone.
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  8. #8
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    {a}_{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}<br />
=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{n}  }\frac{1}{k}<br />
<\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{{k}^{3/2}}

    The last expression tends to zero, since it is the product of a convergence sequence (the series) and an infinitesimal one. This proves that a_n tends to zero
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  9. #9
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    This is a subtopic of the problem of sequence of means.
    Suppose that (a_n\ge 0) is a sequence define S_n  = \sum\limits_{k = 1}^n {a_k } \,\& \,M_n  = \frac{{S_n }}{n}.
    Theorem: If (a_n) monotonic then (M_n) is monotonic.
    Theorem: If (a_n)\to L then (M_n)\to L.
    Last edited by Plato; April 24th 2010 at 01:16 PM.
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